+
BCH 312
Experiment (3)
Titration of a
weak acid with
strong base
+
Objectives
1) 
To study titration curves.
2) 
To determine the pKa value of a weak acid.
3) 
To reinforce the understanding of buffers.
+ Weak Acid
n 
Weak acids or bases do not dissociate completely, therefore
an equilibrium expression with Ka must be used.
n 
The Ka is a quantitative measure of the strength of an acid
in solution.
n 
Ka is usually expressed as pKa.
n 
n 
n 
pKa = -log Ka
As an acid gets weaker, its Ka gets smaller and pka gets
larger and vise versa, for example:
n 
HCl is a strong acid , it has a large value of Ka (low pKa)
n 
CH3COOH is a weak acid , it has low value of Ka (high pKa)
pKa values of weak acids can be determined mathematically
or practically by the use of titration curves.
+
Titration curves
n 
There are many uses of titration, one of them is to indicate the
pKa value of the weak acid by using the titration curve.
n 
Titration Curves are produced by monitoring the pH of a
given volume of a sample solution after successive addition
of acid or alkali.
n 
The curves are usually plots of pH against the volume of
titrant added.
+
n 
How to calculate the pH at each point of
the titration curve
[1] Before any addition of strong
base the (starting point):
n 
The weak acid is in the full protonation
form [CH3COOH] (electron donor) .
n 
In this point pH of weak acid < pKa
We can calculate the pH from:
n 
n 
n 
pH=(𝒑𝑲𝒂+𝑷[𝑯𝑨]) /𝟐
[2] When certain amount of strong
base added (any point before the
middle of titration):
n 
The weak acid is starting to dissociate ,
[CH3COOH]>[CH3COO-] ( Donor >
Acceptor).
In this point the pH of weak acid < pKa
n 
We can calculate the pH from:
n 
n 
pH=pKa+ log[A-]/[HA]
1
2
+Continue
n 
[3] At middle of titration:
n  [CH3COOH]=[CH3COO-] (Donor=Acceptor) ,
n  PH=Pka .
n  The component of weak acid work as a Buffer (A
solution that can resistant the change of PH) .
n  Buffer capacity= pKa ± 1
Note: pKa is defined as the pH value at middle of titration at
which they will be [donor]=[acceptor].
n 
[4] any point after mid of titration:
n  [CH3COOH]< [CH3COO-] , (Donor< Acceptor) .
n  In this point the pH > pKa
n  We can calculate the pH from:
n 
n 
pH= pKa + log[A-]/[HA]
[5] At the end point
n  The weak acid is fully dissociated [CH3COO-] (electron
acceptor) .
n  pH of weak acid >pka
n  Approximatly, all the solution contains CH3COONa so
we first must calculate pOH, then the pH:
n  pOH= (pKb+p[A-]) /2
n  pH= pKw- pOH
5
3
4
+
Notes:
The pH calculated by different way :
n 
[1] At starting point
pH= (pKa+p[HA]) /2
n 
[2] At any point within the curve (after , in or after middle titration)
pH=pKa+ log[A-]/[HA]
n 
[3] At end point
(Henderson-Hasselbalch equation)
pOH=(pKb+p[A-]) /2
pH=pKw – pOH
Henderson-Hasselbalch equation is an equation that is often used to :
n 
To calculate the pH of the buffer
n 
Used in buffer preparation.
n 
To calculate the pH in any point within the titration curve (Except starting and
ending point)
Example
+
n 
Determine the pH value of 500 ml weak acid (0.1M) , titrated with
0.1M KOH (Pka=5) ?
After addition 100 ml , 250 ml , 375 and 500 ml of KOH??
[1] pH after the addition of 100 ml of KOH?
n 
pH=pKa+ logA- /HA
HA + KOH à KA + H2O
n 
No. of moles of HA remaining = No. of moles of HA originally –No. of
moles of KOH
n 
No. of moles of KOH = M x V (in L)= 0.1 X 0.1 = 0.01 moles = No. of
moles of A-
n 
No. of moles of HA originally = M x V (in L) = 0.1 X 0.5 = 0.05 moles
n 
No. of moles of HA remaining = 0.05 – 0.01 = 0.04 moles
n 
pH = 5 + log 0.01 / 0.04
n 
pH = 4.4
+
[2] pH after the addition of 250 ml of KOH??
n 
No. of moles of HA remaining = No. of moles of HA originally –
No. of moles of KOH
n 
No. of moles of KOH = 0.1 X 0.25 = 0.025 moles = No. of
moles of A-
n 
No. of moles of HA originally = 0.1 X 0.5 = 0.05 moles
n 
No. of moles of HA remaining = 0.05 – 0.025 = 0.025 mole
n 
pH = 5 + log 0.025/0.025
n 
pH= 5 = Pka (at mid point , The component of weak acid work
as a Buffer , has a buffering capacity 5 ± 1 )
+
[3] pH after the addition of 375 ml of KOH??
n 
No. of moles of HA remaining = No. of moles of HA originally –
No. of moles of KOH
n 
No. of moles of KOH = 0.1 X 0.375 = 0.0375 moles = No. of
moles of A-
n 
No. of moles of HA originally = = 0.1 X 0.5 = 0.05 moles
n 
No. of moles of HA remaining= 0.05 – 0.0375 = 0.0125 mole
n 
pH = 5 + log 0.0375/0.0125
n 
pH= 5.48
+n  [4] pH after
the addition of 500 ml of KOH??
Note: 500 ml the same volume of weak acid that mean the all weak acid as [CH3COO]
n 
pOH = (pKb + p[A-]) /2
pKb= pKw-pKa !pKb =14-5 = 9
n 
p[A-]= - log [A-] à [A-]=??
n 
No of moles of KOH = 0.1 X 0.5 = 0.05 moles = No. of moles of A-
n 
Total volume = 500 + 500 = 1000 = 1L
n 
[A-] = 0.05/1 =0.05 M
n 
p[A-]= - log 0.05 = 1.3
n 
pOH= (9 +1.3 ) / 2 = 5.15
n 
pH=pKw-pOH
n 
PH = 14 – 5.15 = 8.85 (at end point)
+
Method
n 
You are provided with 10 ml of a 0.1M CH3COOH weak acid
solution, titrate it with 0.1m NaOH adding the base drop wise
mixing, and recording the pH after each 0.5 ml NaOH added
until you reach a pH=10.
Volume of 0.1 NaOH (ml)
0
0.5
1
1.5
pH
+
Discussion:
1- Plot a Curve of pH versus ml of NaOH added, determine the
pKa value from the curve.
2- Calculate the pH of the weak acid HA solution after the
addition of 3ml, 5ml, and 10ml of NaOH.
3- Compare your calculated pH values with those obtained
from Curve.
4- At what pH-range did the acid show buffering behavior?
What are the chemical species at that region, what are their
proportions?